When $M$ is a type I von Neumann algebra on a Hilbert space, then the commutant $M'$ is also type I.
The conclusion is also true for type II and type III von Neumann algebras.
My question is as following:
If $M$ is a finite(semi-finite,respectively) von Neumann algebra, can we conclude that $M'$ is also finite(semi-finite,respectively)?
Best Answer
This is answered in all von Neumann algebra texts.
The "major" type (I,II,III) is preserved in the commutant; so "semifinite" is preserved (note that the commutant of a direct sum/direct integral is the direct sum/integral of the commutants). The "minor" type (I$_n$, II$_1$, II$_\infty$) is not preserved. Indeed, it is very easy to construct examples. For instance when $M_6(\mathbb C)$ is seen as $M_2(\mathbb C)\otimes M_3(\mathbb C)$, we have $M=M_2(\mathbb C)\otimes1$, $N=1\otimes M_3(\mathbb C)$ two subalgebras with $M$ type I$_2$, $N$ type I$_3$, and $M'=N$.
The same happens with any finite algebra. If $M\subset B(H)$ is finite, then you can always consider $M\otimes 1\subset B(H\otimes \ell^2(\mathbb N))$ and then $1\otimes B(\ell^2(\mathbb N)) \subset (M\otimes 1)'$, showing that $M\otimes 1$ cannot be finite. In this situation the commutant of a II$_1$ algebra can be II$_\infty$ and the commutant of a I$_n$ can be I$_\infty$.